Joint density $X^2+Y^2$ Let's say we have a point $(x,y)$ in the unit circle.
I've read (without proof :( ) that the joint density of $z$, where $z^2=x^2+y^2$, is:
$$f_{X,Y}(x,y) = \begin{cases}\frac{2}{\pi}(x^2+y^2)&\mathrm{\ if \ } x^2+y^2\le1\\ 0&\mathrm{\ otherwise\ }\end{cases}$$
Can someone show me why? 
 A: What you say is (almost?) true if you say that you're random point in the circle is uniformly distributed.  That means the probability of falling within any region is proportional to the region's area.  The area of a circle centered at $(0,0)$ whose radius is $z$ is $\pi z^2$.  This is $\pi(x^2+y^2)$.  The probability of falling within that circle is therefore $\pi(x^2+y^2)$ divided by the area of the whole unit circle, which is $\pi$.  But notice you're talking about the joint density of $X$ and $Y$ (not to be confused with $x$ and $y$, not the density of $z$.  The cumulative probability distribution function (not the density function) of $Z=\sqrt{Z^2+Y^2}$ is therefore
$$
F_Z(z) = \begin{cases} z^2 & \text{if }0\le z\le1, \\  0 & \text{if }z<0, \\  1 & \text{if }z>1.  \end{cases}
$$
The density function is therefore
$$
f_Z(z) = \begin{cases} 2z & \text{if }0\le z\le1, \\ 0 & \text{otherwise.} \end{cases}
$$
The joint density of the coordinates is
$$
f_{X,Y}(x,y) = \begin{cases} 1/\pi & \text{if }x^2+y^2\le1, \\ 0&\text{otherwise}. \end{cases}
$$
Here you see the reason why the word "uniform" is used: the thing that's "uniform" is the density function, in this cases a constant $1/\pi$.
